Minimum Operations to Make a Special Number - O(n) - Two Pointers [JS]

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Description 

Solution: Two Pointers

Find the minimum operations to make num end with one of ['00', '25', '50', '75'].
Use two pointers and iterate through num from right to left, greedily matching the end digits.

Special case: We can also turn num into 0, which will take n - 1 operations if num contains '0', otherwise n operations.

Time Complexity: O(5n)
Space Complexity: O(1)

var minimumOperations = function(num) {
  let possibleEndDigits = ['00', '25', '50', '75'];
  let minOperations = num.includes('0') ? num.length - 1 : num.length; // turn number into 0
  for (let endDigits of possibleEndDigits) {
    minOperations = Math.min(minOperations, operationsToEndWith(num, endDigits));
  }
  return minOperations;
};

function operationsToEndWith(num, endDigits) {
  let n = num.length, operations = 0, j = endDigits.length - 1;
  for (let i = n - 1; i >= 0; i--) {
    if (num[i] === endDigits[j]) {
      j--;
      if (j === -1) return operations;
    } else {
      operations++;
    }
  }
  return n;
}
Location: Auckland, New Zealand

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